STOICHIOMETRY HONORS CHEMISTRY

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Transcription:

STOICHIOMETRY HONORS CHEMISTRY

MOLE RATIO A mole ratio is the ratio of coefficients used to compare amounts of reactants and products. 1 ZnCl 2 (aq) + 2 NaOH (aq) 1 Zn(OH) 2 (aq) + 2 NaCl (aq) What is the mole ratio of NaOH to NaCl? 2:2 or 2 moles NaOH: 2 moles NaCl What is the mole ratio of ZnCl 2 to NaCl? 1:2 or 1 mole ZnCl 2 1 mole NaCl note that the coefficient for ZnCl 2 is listed first, and the coefficient for NaCl is listed second because that is the order of the substances in the question.

MOLE TO MOLE 1 step conversion A mole to mole 1 step conversions uses the ratio of coefficients to compare amounts of reactants and products. 4 FeS + 7 O 2 2 Fe 2 O 3 + 4 SO QUESTION 1: What amount in moles of O 2 will be required to produce 6.4 moles of SO? unit wanted as answer 6.4 moles SO x 7 moles O 2 = (6.4 moles SO)(7 moles O 2 ) = 11.2 mole O 2 4 moles SO 4 moles SO given in problem given unit cancel out moles SO

GRAM TO GRAM 3 step conversion 4 FeS + 7 O 2 2 Fe 2 O 3 + 4 SO QUESTION 2: If 15.88 grams O 2 of oxygen react, what mass of SO will be made? molar mass SO = 32 + 16 15.88 grams O 2 x 1 mole O 2 x 4 moles SO x 48 grams SO = (15.88 grams O 2 )(1 mole O 2 )(4 moles SO)(48 g SO) = 13.20 g SO 32.998 g O 2 7 moles O 2 1 mole SO (32.998 g O 2 )(7 moles O 2 )(1 mole SO) Molar mass O 2 = 15.999x2 15.88 x 1 x 4 x 48 32.998 7 1 = 13.20 g SO Observe the pattern of units: Grams A x 1 moles A x moles B x grams B = grams B molar mass A moles A molar mass B

GRAM TO MOLE 2 step conversion 4 FeS + 7 O 2 2 Fe 2 O 3 + 4 SO QUESTION 3: If 15.88 grams O 2 of oxygen react, what amount in moles of SO will be made? 15.88 grams O 2 x 1 mole O 2 x 4 moles SO = (15.88 grams O 2 )(1 mole O 2 )(4 moles SO) = 32.998 g O 2 7 moles O 2 (32.998 g O 2 )(7 moles O 2 ) 15.88 X 1X 4 32.998 7 = 0.27 moles SO The gram to mole 2 step conversion is the same as the 3 step conversion, but excludes the third step, leaving the value in moles.

MOLE TO GRAM 2 step conversion 4 FeS + 7 O 2 2 Fe 2 O 3 + 4 SO QUESTION 4: If 7.3 moles O 2 of oxygen react, what mass of SO will be made? 7.3 mol O 2 x 4 moles SO x 48 grams SO = 7.3 mole O 2 )(4 moles SO)(48 g SO) = 200.23 g SO 7 moles O 2 1 mole SO 7 moles O 2 )(1 mole SO) start with mole ratio The mole to gram is a 2 step conversion. The math is the same as the 3 step conversion, but excludes the first step of converting grams to moles because the starting value is already in moles.

VOCABULARY LIMITING REACTANT the reactant that runs out first. The limiting reactant controls how much product that can be made because when the reactant runs out the reaction stops. EXCESS REACTANT-the reactant there is plenty of. The reactant there is leftovers of because all of the excess reactant isn t used. THEORETICAL YIELD the maximum amount of product that will be made from a chemical reaction. The theoretical yield is calculated from a grams to grams (3 step) calculation. The theoretical yield is dependent on the limiting reactant. ACTUAL YIELD the mass of product made experimentally in the lab setting. % YIELD is a mathematical comparison of how much product was made compared to how much should have been made. % yield = ACTUAL YIELD (100) THEORETIAL YIELD

Calculate the percentage yield of NaBr if 20 grams NaI and 12 g Br 2 react and form 12.3 g NaBr in the lab? 2 NaI + Br 2 2 NaBr + I 2 Start with 3 step converting grams of reactant to grams of product. 20 grams NaI x 1 mole NaI x 2 moles NaBr x 102.89 grams NaBr = 13.7 g NaBr 149.89 g NaI 2 mole NaI 1 mole NaBr limiting reactant molar mass NaI = 22.990+126.904 molar mass NaBr = 22.990+79.904 12 grams Br 2 x 1 mole Br 2 I x 2 moles NaBr x 102.89 grams NaBr = 15.45 g NaBr 159.81 g Br 2 1 mole Br 2 1 mole NaBr excess reactant molar mass Br 2 = 79.904 x 2 molar mass NaBr = 22.990+79.904 Identify the smaller value as theoretical yield. 13.7 g NaBr is the smaller value, therefore the theoretical yield. 15.45 grams of NaBr will never be made because once 13.7 g NaBr is made the reaction runs out of NaI, therefore the reaction stops producing NaBr product. NaI is known as the limiting reactant. Solve for percent yield. ACTUAL YIELD (100) = 12.3 grams NaBr (100) = 89.8 % THEORETICAL YIELD 13.7 grams NaBr Hint: actual yield will always be given in problem, theoretical yield will always be calculated with a 3 step grams to grams conversion

IDENTIFY THE LIMITING REACTANT WHEN 8.7 g Na REACTS WITH 9.3 g FeCl 3? 3 Na + 1 FeBr 3 3 NaBr + Fe 8.7 grams Na x 1 mole Na x 3 moles NaBr x 102.89 grams NaBr = 38.9g NaBr 22.990 g Na 3 mole Na 1 mole NaBr 9.3 grams FeCl 3 x 1 mole Na x 3 moles NaBr x 102.89 grams NaBr =17.7 g NaBr 161.83 g FeCl 3 1 mole FeCl 3 1 mole NaBr FeCl 3 is the limiting reactant because it makes a smaller amount of product (NaBr). 17.7 grams NaBr is less then 38.9 g NaBr

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